Question
Answer
$$\dfrac{\sqrt{5}-\sqrt{7}}{2\sqrt{3}+2}=\dfrac{\sqrt{15}-\sqrt{5}-\sqrt{21}+\sqrt{7}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{5}-\sqrt{7}}{2\sqrt{3}+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{5}-\sqrt{7}}{2\sqrt{3}+2}\frac{2\sqrt{3}-2}{2\sqrt{3}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{15}-2\sqrt{5}-2\sqrt{21}+2\sqrt{7}}{12-4\sqrt{3}+4\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{15}-2\sqrt{5}-2\sqrt{21}+2\sqrt{7}}{8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{15}-\sqrt{5}-\sqrt{21}+\sqrt{7}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3}-2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{5}- \sqrt{7}\right) } \cdot \left( 2 \sqrt{3}-2\right) = \color{blue}{ \sqrt{5}} \cdot 2 \sqrt{3}+\color{blue}{ \sqrt{5}} \cdot-2\color{blue}{- \sqrt{7}} \cdot 2 \sqrt{3}\color{blue}{- \sqrt{7}} \cdot-2 = \\ = 2 \sqrt{15}- 2 \sqrt{5}- 2 \sqrt{21} + 2 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3} + 2\right) } \cdot \left( 2 \sqrt{3}-2\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot-2+\color{blue}{2} \cdot 2 \sqrt{3}+\color{blue}{2} \cdot-2 = \\ = 12- 4 \sqrt{3} + 4 \sqrt{3}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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